ZEROS OF POLYNOMIAL FUNCTIONS ALL I HAVE TO KNOW ABOUT POLYNOMIAL FUNCTIONS
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1 ZEROS OF POLYNOMIAL FUNCTIONS ALL I HAVE TO KNOW ABOUT POLYNOMIAL FUNCTIONS
2 TOOLS IN FINDING ZEROS OF POLYNOMIAL FUNCTIONS Synthetic Division and Remainder Theorem (Compressed Synthetic Division) Fundamental Theorem of Algebra Factor Theorem Rational Roots Theorem Upper Bound and Lower Bound Theorems Conjugate Roots Theorem Irrational Roots Theorem Intermediate Value Theorem
3 Fundamental Theorem of Algebra A polynomial function of degree n has at most n zeros. Every polynomial function of degree n 1 has exactly n zeros, where a zero of multiplicity m is counted m times.
4 Compressed Synthetic Division with Rational Roots Theorem Given: P(x) = 2x 4-9x 3 + 7x 2 + 9x - 9 a 0 = -9 a n = 2 c (factors): ±1, ±3, ±9 RRT d (factors): ±1, ±2 Possible rational roots: c d : ±1, ± 1 2, ±3, ± 3 2, ±9, ± 9 2
5 Given: P(x) = 2x 4-9x 3 + 7x 2 + 9x - 9 Possible rational roots: c d : ±1, ± 1 2, ±3, ± 3 2, ±9, ± 9 2 x P(x) Lower Bound / is a zero 1 is a zero is a zero Upper Bound Change of signs in P(x): Intermediate Value Theorem Compressed Synthetic
6 RATIONAL ROOTS THEOREM Let P(x) = a n x n + a n-1 x n a 1 x + a 0 where all the coefficients are integers. Consider a rational number denoted by c, where c and d are relatively prime. For c d d to be a zero of function P, c must be a factor of a 0 (the constant) and d must be a factor of a n (the leading coefficient) Example 1
7 Upper Bound Definition: A number on the positive x-axis to the right of which the graph does not change the direction anymore is called an upper bound. Upper Bound Theorem: Example 1 Consider a polynomial function P. Let r be a positive number. If synthetic division is used to divide P(x) by x r and a third row of numbers having the same sign is obtained, then r is an upper bound for the zeros of P.
8 Lower Bound Definition: A number on the negative x-axis to the left of which the graph does not change the direction anymore is called a lower bound. Lower Bound Theorem: Example 1 Consider a polynomial function P. Let s be a negative number. If synthetic division is used to divide P(x) by x s and a third row of numbers whose signs are alternating is obtained, then s is a lower bound for the zeros of P.
9 Intermediate Value Theorem For any polynomial p with real coefficients, suppose that p(a) p(b) and a < b. Also, suppose that the function values have opposite signs. Then p has a real zero between a and b (b, P(b)) P(b) > 0 or positive P(a) < 0 or negative (a, P(a)) Therefore, the function has a real zero between a and b. Example 1
10 Given: P(x) = 2x 4-9x 3 + 7x 2 + 9x - 9 Possible rational roots: c d : ±1, ± 1 2, ±3, ± 3 2, ±9, ± 9 2 x P(x) is a zero Factors of P(x) are: P(x) = (x + 1) (x -1) (2x 2 9x + 9) P(x) = (x + 1) (x -1) (2x 3) (x 3) The degree of this quotient is 2 nd degree Therefore, zeros of P are -1, 1, 3/2, and 3 1 is a zero Use this as new dividend This is factorable: (2x 3)(x 3)
11 Example 2. Given: p(x) = x 3 - x 2-3x + 3 Possible rational roots: 1, 3 x P(x) Example is a zero Factors of P(x) are: P(x) = (x 1) (x 2 3) Zeros of P are 1, 3,- 3 Irrational Roots Theorem Not factorable but can be solved: x 2 3 = 0 x 2 = 3 x = ± 3
12 Example 3. Given: p(x) = x 3 - x 2 + 2x - 2 Possible rational roots: 1, 2 x P(x) Exercise 1 1 is a zero Factors of P(x) are: P(x) = (x 1) (x 2 + 2) Zeros of P are 1, i 2,-i 2 Conjugate Roots Theorem Not factorable but can be solved: x = 0 x 2 = -2 x = ±i 2
13 Irrational Roots Theorem Suppose P(x) is a polynomial with rational coefficients and of degree greater than or equal to 1. Then either of the following is a root, so is the other: a + c b,a - c b where a, b, and c are rational numbers and b is irrational. Example 2
14 Conjugate Roots Theorem If a polynomial P(x) of degree greater than or equal to 1 with real coefficients has a complex number a + bi as a root, then its conjugate a bi is also a root. (Complex roots occur in conjugate pairs.) Example 3
15 Examle 4. Given: P(x) = 3x 5 + 2x 4 22x 3 17x x - 4 How many zeros does function P have? 5 zeros What are the possible rational zeros of P? c d : ±1, ± 1 3, ±2, ± 2 3, ±4, ± 4 3 x P(x) By Fundamental Theorem of Algebra, is it possible to use -2 as a divisor? By Remainder Thrm: -2 is a zero of P By IVT: There are zeros between 0 and 4
16 Given: P(x) = 3x 5 + 2x 4 22x 3 17x x - 4 Possible rational zeros: c d : ±1, ± 1 3, ±2, ± 2 3, ±4, ± 4 3 x P(x) Therefore, 1/3 is another zero of P. The factors of P(x) are: P(x) = (x +2)(x+2) (x 1/3)(3x 2 9x +3) -2 is a zero with multiplicity 2 Since there is a zero between 0 and 3, then we can try the fractions on possible rational zeros
17 Given: P(x) = 3x 5 + 2x 4 22x 3 17x x - 4 x P(x) The factors of P(x) are: P(x) = (x +2)(x+2) (x 1/3)(3x 2 9x +3) P(x) = (x +2)(x+2) (x 1/3)3(x 2 3x +1) P(x) = (x +2)(x+2) (3x 1)(x 2 3x +1) The coefficients of the factors must be integral Factor out 3 Multiply 3 to (x 1/3) to make the coefficient integral.
18 Given: P(x) = 3x 5 + 2x 4 22x 3 17x x - 4 x P(x) P(x) = (x +2)(x+2) (3x 1)(x 2 3x +1) The zeros of P are -2 ( Multiplicity 2), 1 3, Use quadratic formula in the quadratic factor, 3-5 2
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